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The electric field of a signed point electric charge in vacuum follows Coulomb's law:For the conducting plane, the method of images replaces each charge $Q_j$ at height $h_j$ by an image charge $-Q_j$ at height $-h_j$. The real and image electric fields have cancelling tangential components and equal normal components at the surface. Thus the electric field at a conducting plane from axial point charges, just above the point directly below them, isNumerically,The surface value means the air-side limit. Inside the ideal conductor the electrostatic electric field is zero; its discontinuity across the surface is supported by induced surface charge. Omitting the image charges would give half the external answer.

The method of images replaces electrical conductors by fictitious image charges outside the physical region. Choose the image charges so that the physical electric charges and their images produce the prescribed constant electric potential on every conducting boundary. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in the charge-containing region. Several conducting planes can require several, or infinitely many, image charges; a single reflected image is sufficient for one infinite plane.

Put the plane at $z=0$ and the point charge at $(0,0,d)$. For $z>0$, use an image charge $-q$ at $(0,0,-d)$:This vanishes on the plane. An arbitrary constant can be added if a different reference electric potential is chosen. Excluding the point charge's own electric field, Coulomb's law givesThe separation from the image charge is $2d$, and the force is attractive for either sign of $q$. No imposed external electric field is included.

Electrical contact equalizes electric potential and preserves total electric charge. The usual isolated-sphere approximation assigns the two conducting spheres the capacitances $4\pi {ϵ}_{0}R$ and $8\pi {ϵ}_{0}R$. It then givesThese are accurate when the conducting spheres are far apart and joined by a thin wire. Literal touching changes the answer: each conducting sphere polarizes the other, so its electric potential is not its isolated value. The source does not specify an approximation which would remove this effect.

For completeness, the exact result for ideal externally tangent conducting spheres can be obtained by the method of images. Let their radii be $a,b$, their centre separation $D=a+b$, and their common electric potential $V$. Begin with central image charges $4\pi {ϵ}_{0}Va$ and $4\pi {ϵ}_{0}Vb$. Reflect each external image charge $q^{\prime }$ at centre-distance $s$ into the other conducting sphere of radius $a$: the new image charge is $-a{q}^{\prime }/s$ at centre-distance $a^2/s$ toward the original. Repeat, alternately, in the two conducting spheres. Pairing successive opposite-sign terms gives the convergent net electric chargesEach reflection cancels a nonconstant surface electric potential; the remaining central term fixes $V$. The total internal image charge equals the exterior electric flux divided by $1/{ϵ}_0$, so these sums give the physical net electric charges.

For $a=R,b=2R$, multiply both dimensionless sums by the common factor $4\pi {ϵ}_{0}V(2R/3)$. Their values follow by writing reciprocals as integrals and setting $t=u^3$:Dividing by $S_a+S_b$ enforces conservation of electric charge and yieldsThis concrete touching-sphere result disproves applying the isolated $1:2$ sharing rule to literal contact.

Take the grounded plane as $z=0$. The method of images replaces it, in $z>0$, by an image charge $-q$ at $z=-d$ and an image charge $+q$ at $z=-2d$. These make the electric potential zero on the plane; the uniqueness theorem for the electrostatic potential establishes the solution in the physical half-space.

At $z=3d/2$, both real point charges contribute upward electric fields, each of magnitude $kq/(d/2{)}^2$. The nearer image charge contributes downward and the farther one upward. HenceThis is upward for $q>0$; retaining signed $q$ automatically handles the opposite sign. No self-electric field is involved because the evaluation point is distinct from every point charge.

For image charges in a sixty-degree conducting wedge, use polar coordinates in a cross-section normal to the planes' intersection, putting the physical wedge at $0<\theta <\pi /3$ and the real point charge at $(r,\theta )=(d,\pi /6)$. Repeated reflection in the two grounded faces produces five image charges, all at radius $d$ and the same coordinate along the intersection:Equivalently, the positive copies are at $\theta =\pi /6+2k\pi /3$ and the negative copies at $\theta =-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the original point charge lies inside the physical wedge.

For any observation point on either face, reflection pairs every positive point charge with an equidistant negative one. Their Coulomb potentials cancel, so both grounded faces have zero electric potential. The electric potential also tends to zero at infinity. The uniqueness theorem for the electrostatic potential then identifies this method of images construction with the physical solution.

Let ${\hat{\mathbf{e}}}_r$ point from the intersection to the real point charge. Reflection symmetry cancels the component of electrostatic force perpendicular to the wedge bisector. An image charge $q_j$ at angular separation $\Delta$ from the real point charge gives radial forceThe two adjacent negative image charges have $\Delta =\pm 60^{\circ }$, distance $d$, and combined radial force $-k{q}^2/d^2$. The two positive image charges have $\Delta =\pm 120^{\circ }$, distance $\sqrt{3}d$, and combined radial force $+k{q}^2/(\sqrt{3}{d}^2)$. The opposite negative image charge is at distance $2d$ and contributes $-k{q}^2/(4d^2)$. ThusThe electrostatic force is towards the intersection, independently of the sign of $q$. One sums the five image charge electric fields and multiplies by $q$; the factor $1/2$ used in induced-charge electrostatic energy must not be inserted into this direct force calculation.

Take the grounded electrical conductor to occupy $z\le 0$ and approximate the negligible-radius charged sphere by a point charge $Q$ at ${\mathbf{r}}_{+}=D\hat{\mathbf{z}}$. Put a fictitious image charge $-Q$ at ${\mathbf{r}}_{-}=-D\hat{\mathbf{z}}$. The constructed electric potential is zero on $z=0$, has the correct point charge singularity, and vanishes at infinity. The uniqueness theorem for the electrostatic potential makes it the physical electric potential for $z>0$:Taking its negative gradient gives the requested electric field outside the small sphere:On the plane at cylindrical radius $r$, the tangential components cancel and the normal component isThe electrostatic boundary conditions at a conductor, with outward unit normal $+\hat{\mathbf{z}}$ and zero interior electric field, give the induced surface charge densityFinally integrate over annuli of area $2\pi rdr$:The image charge is a computational device, not a real charge below the plane. For a finite conducting sphere, its induced polarization would change the point-source approximation; the stated negligible radius is essential.

Set the grounded plane at $z=0$ and the real point charge at $(0,0,h)$. The image charge $-q$ at $(0,0,-h)$ makes the electrostatic potential vanish throughout the plane. The uniqueness theorem for the electrostatic potential then identifies the physical solution in $z>0$:The normal electric field is $E_z=-{\partial }_z\Phi$. Differentiation givesand thereforeFor $q>0$ the electric field points into the electrical conductor, consistently with negative induced surface charge density.

Apply the superposition principle to the result of part (a). If $F(a,h)=-qh/[2\pi {ϵ}_0(a^2+h^2{)}^{3/2}]$, then the exact normal electric field from the two real point charges and their image charges isFor $d\ll h$, Taylor's theorem gives $E_z=-d{\partial }_{h}F+O(d^2)$. Sincethe leading electric dipole result isThe electric dipole moment is $\mathbf{p}=-qd\hat{\mathbf{z}}$. The circle $a=\sqrt{2}h$ is the zero of the leading normal electric field and hence of the induced surface charge density. Inside it, induced surface charge density is negative; outside it, positive. The tangential electric field is zero everywhere on the grounded electrical conductor, so the leading full electric field vanishes on that circle. The corresponding dipole-induced charge on a grounded plane integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.

For finite $d$, $a=\sqrt{2}h$ is only the leading-order zero. With $u=h^{2/3}$ and $v=(h+d{)}^{2/3}$, the exact zero satisfies $a^2=uv(u+v)$, giving $a=\sqrt{2}(h+d/2)+O(d^2/h)$. For example, inserting any $d>0$ in the exact expression at $a=\sqrt{2}h$ does not give zero. This is a controlled electric dipole approximation, not an exact finite-separation identity.

The image charge construction replaces conductor-induced charge by fictitious sources outside the physical domain. Match the conductor potentials and the far-field condition; uniqueness then fixes the potential in the physical region. For a grounded plane, reflect a charge and reverse its sign, giving attraction $q^2/(16\pi {ϵ}_0{h}^2)$. For a grounded sphere of radius $a$, a charge at distance $R>a$ has image $-qa/R$ at radial distance $a^2/R$. These are examples with prescribed zero potential, not arbitrary isolated conductors.^[ref]

Use horizontal $x$ and upward $y$ in the upper right-angle wedge. The sheets meet electrically, so take their common potential as zero and assume no externally imposed field. For a real charge at $(0,y)$, the image charges in a right-angle conducting wedge are $-q$ at $(y,0)$ and $(-y,0)$, and $+q$ at $(0,-y)$. The first two contributions have cancelling horizontal components. With $K_e=1/(4\pi {ϵ}_0)$,Thus the force magnitude at the original height is $\beta /y_0^2$, directed down toward the intersection. The sphere is treated as a point charge, with size much smaller than its distances from the sheets.

Since gravity is already balanced in the uncharged equilibrium, the charged vertical equilibrium satisfies $k(y_e-y_0)+\beta /y_e^2=0$. For $\beta /(k{y}_0^3)\ll 1$, $y_e\simeq y_0-\beta /(k{y}_0^2)$. Differentiating the force givesThe electrostatic softening above a conducting wedge reduces the restoring force; the small-charge vertical equilibrium remains stable.

For the lateral stability of a spring-supported charge in a wedge, the off-axis images of $(x,y)$ are $-q$ at $(y,x)$ and $(-y,-x)$ and $+q$ at $(-x,-y)$. Expansion near the bisector givesSo the electrostatic force is laterally destabilizing: moving toward either sheet strengthens its attraction. In the simplified model where the spring supplies only a vertical force, the overall equilibrium is a saddle and is unstable.

A three-dimensional spring suspension also has geometrical transverse stiffness. If its charged-equilibrium tension is $T$ and length is $\ell$, a small sideways displacement gives spring force $-(T/\ell )x$, with $T=mg+\beta /y_e^2$. The full stability conditions in this suspension model areMotion parallel to the intersection line has no electrostatic stiffness in the ideal infinite-wedge geometry and is stabilized by the positive spring tension. Thus lateral electrostatic destabilization alone does not prove instability of a real hanging spring. At fixed finite support geometry, a sufficiently small charge preserves stability because $D_x=O(q^2)$ while $T/\ell \to mg/\ell >0$. The given vertical-displacement approximation does not specify the attachment length or by itself establish the transverse inequality for an arbitrary charge. The general criterion above states what can be concluded; the saddle verdict applies to the distinct idealization with zero lateral support stiffness.

An electrostatic field calculation begins by specifying the charge information, material response and boundary conditions. In vacuum, prescribed volume and surface electric charge densities can be superposed directly using Coulomb's law. For a volume distribution,Differentiating this electric potential gives the electric field. This is useful for known source distributions, but in the presence of conductors the induced surface electric charge is usually an unknown rather than an independently specified input.

Gauss's law is especially effective when symmetry fixes the direction and magnitude of the electric field on a suitable closed surface. A spherical charge distribution gives a radial field, an infinitely long uniform line gives a cylindrical field, and a uniform infinite sheet gives a planar field. In an arbitrary geometry, the flux law remains true but does not alone determine the pointwise electric field. A multipole expansion is another useful approximation: far from a localized charge distribution, successive monopole, dipole and higher terms separate the effects of total charge and its spatial structure.

For a boundary-value formulation, use the electrostatic equationsThey imply Poisson equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$, or Laplace equation where $\rho =0$. A conductor in electrostatic equilibrium is an equipotential with zero internal electric field. Its exterior normal electric field fixes the surface electric charge through $E_n=\sigma /{ϵ}_0$. One must specify whether each conductor is grounded, maintained at a prescribed voltage, or isolated with prescribed total electric charge.

Uniqueness of Poisson equation justifies methods that construct a convenient candidate solution. With fixed boundary electric potential and suitable behaviour at infinity, a solution satisfying the field equation is the solution. With normal-derivative boundary conditions, the source and flux must be compatible, and the electric potential is determined only up to an additive constant.

For example, a point charge $q$ a distance $h$ above a grounded infinite plane has the same upper-half-space electric potential as $q$ and a mirrored image charge $-q$ below it. Their potentials cancel on the plane. The image is a mathematical source in the excluded region, not a second physical charge. Separation of variables similarly constructs solutions suited to rectangular, cylindrical or spherical boundaries; matching coefficients enforces the prescribed boundary data.

In dielectrics, the distinction between total and free charge is essential. In a linear isotropic material,Only where permittivity is constant may this be reduced to ${\nabla }^2\varphi =-{\rho }_f/ϵ$. At an ordinary interface, $\varphi$ and the tangential electric field are continuous, while the jump in normal electric displacement field is the free surface charge. Applying Gauss's law to total charge with ${ϵ}_0$, or its displacement-field form to free charge, gives consistent descriptions; mixing their sources and permittivities does not.

A useful interface example is the image charge at a dielectric interface. If $q$ lies in medium 1, the source-side potential is represented bywhere $r_1$ is distance to the real charge and $r_2$ to its mirrored position. The other-side potential can be written ${\varphi }_2=q_t/(4\pi {ϵ}_2{r}_1)$, where $q_t=2{ϵ}_{2}q/({ϵ}_1+{ϵ}_2)$. Continuity of $\varphi$ and of normal $\mathbf{D}$ fixes both coefficients. The equal-permittivity limit gives no image; the conductor limit ${ϵ}_2\to \infty$ gives $q^{\prime }\to -q$.

For a sphere of radius $a$ and relative permittivity ${ϵ}_r$ in a uniform external electric field ${\mathbf{E}}_0$, matching the spherical solutions gives a uniform internal electric field and an external induced electric dipole moment:This illustrates how a material boundary modifies an otherwise simple applied electric field.

For complex boundaries or spatially varying materials, finite difference methods or the finite element method solve the same boundary value problem numerically. A sound calculation checks its boundary conditions, charge or flux balance and convergence with mesh refinement. The governing equations, source classification and uniqueness argument remain the organizing principles even when an analytic formula is unavailable.

The original PDF places this complete question before B6 without a visible question number. It is included here under a descriptive heading.

Why images work. In the method of images, place fictitious image charges outside the region where the physical electric field is required. Choose them so that the combined electric potential satisfies the conductor boundary conditions. The Uniqueness of the Dirichlet problem then proves that this construction gives the physical exterior field. The images do not describe the actual surface-charge locations.

Two examples are a charge $q$ above a grounded conducting plane, represented by $-q$ at the mirror point, and the image charge for a grounded conducting sphere used below. In each case the image lies inside the excluded conductor and makes its surface an equipotential of the required value.

Potential of the two charges. By the law of cosines, the distances from the observation point to charges at $x_j$ are $s_j=(r^2+x_j^2-2r{x}_j\cos \theta {)}^{1/2}$. The superposition principle and Coulomb's law therefore give

The grounded sphere. Set $r=a$ and require $s_1=\alpha s_2$ at every polar angle. Matching the coefficients of $\cos \theta$ and the constant terms givesThe useful noncoincident solution isIndeed $s_1=(a/x_2)s_2$ on the sphere, so the two potential terms cancel there. A Gaussian surface enclosing the sphere but not the external charge measures the total image charge in the exterior representation. Thus the physical induced net charge is

An isolated neutral sphere. The grounded image would give nonzero net charge. Add a central image $q_0=+aq/x_2$ to cancel it. A central image changes the surface potential by a constant, preserving the equipotential condition, whereas an arbitrary off-centre image would generally spoil it. The image charges for a neutral conducting sphere giveThe potential is measured relative to zero at infinity for each finite $x_2$.

Fixed applied field and dipole moment. For positive $q$ on the positive $x$ axis, the applied central field is ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{x}_2^2)$. The central image contributes no electric dipole moment, soIts units are coulomb metres. Keeping $E_0$ fixed makes $q/x_2^2$ constant, so this induced moment is independent of the distance.

The distant-charge limit. As $x_2\to \infty$ with $q/x_2^2$ fixed, the applied field becomes uniform over any fixed neighbourhood of the sphere. The neutral pair of opposite images approaches a point dipole of the finite moment above. The external potential contains a growing spatial constant $q/(4\pi {\epsilon }_0{x}_2)$, so the infinity-referenced conductor potential does not have a finite limit. Subtracting this irrelevant constant gives the usual induced charge on a conducting sphere in a uniform electric field; the exterior potential becomesThis has zero potential on the sphere, but a uniform applied field cannot simultaneously have its potential vanish at spatial infinity.

For real point charges $Q$ and $-2Q$ at $(0,b)$ and $(a,b)$ above the plane $y=0$, the image charges are $-Q$ and $+2Q$ at the reflected positions. The image charges enforce constant electric potential on the plane. The physical forces follow by summing ordinary Coulomb's law forces on each real charge; the electrostatic-energy factor of one half for induced charges does not multiply this force. The smaller positive charge can have an upward total force component, even though its own image attracts it downward.